Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

196 D CHAPTER 14 PARTIAL DERIVATIVES
7'(-15 30) ~ f (-10, 30) - f (- 15, 30)
! = - 20 - (- 26) = ~ = 1 2
' 5 5 5 . '
" f{- 20, 30)- f (-15,30) - 33 - {- 26) - 7
h( - 10, 30) ~ , _ 5
= _ 5
. = _ 5
= 1.4. Averaging these values, we estimate
fr( -15, 30) to be approximately 1.3. Thus, when tl1e actual temperature is - 15°C and the wind speed is 30 km/ h, the
apparent temperature rises by about 1.3°C for every degree iliat the actual temperature rises.
S . . 1 I f ( 15 30 ) . f (- 15, 30 +h) - f( -15, 30) . .
tml ar y, v - , = 11m h wh1ch we can approxunate by considering h = 10
l•- 0
and h = - lO: f .., (- 15 , 30 ) ~ f (- 15,40) - f(- 15, 30) = - 27 - (-26) = -
10 10
1 = - 0. 1
,
10
f(-15 20)-f(- 15 30) - 24 - (-26) 2 .
f v{-15, 30) ~ ' _ 10
' = _ 10
= _ 10
= - 0.2. Averaging these values, we estimate
f..,( - 15, 30) to be appr~ximately - 0.15. Thus, when the actual temperature is - 15°C and ilie wind speed is 30 km/ h, ilie
apparent temperature decreases by about 0.15°C for every km/ h that the wind speed increases.
(b) For a fixed wind speed v, the values of the wind-chill index W increase as temperature T increases (look at a column of
the table), so a; is positive. For a fixed temperature 1', the values of W decrease (or remain constant) as v increases
(look at a row of the table), so ~: is negative (or perhaps 0).
(c) For fixed values ofT , the function values f (T, v) appear to become constant (or nearly constant) as v increases, so the
corresponding rate of change is 0 or near 0 as v increases. This suggests that lim ( aw I ov) = 0.
V->00
5. (a) If we start at (1, 2) and mo~e in the positive x-direction, the graph off increases. Thus f,(1, 2) is positive.
(b) If we start at (1, 2) and move in the positive y-direction, tbe graph off decreases. Thus fv(1, 2) is negative.
7. (a) f zx = g, (!,), so f xx is the rate of change of f , in tl1e x -direction. fx is negative at ( - 1, 2) and if we move in the
positive x-direction, the surface becomes less steep. Thus the values of f., are increasing and !:ex(- 1, 2) is positive.
(b) f 1111 is the rate of change of f 11 in tlle y-direction. / 11 is negative at (- 1, 2) and if we move in the positive y-d.irection, the
surface becomes steeper. Thus the values of f 11 are decreasing, and / 1111 ( - 1, 2) is negative.
9. First of all, if we start at the point (3, - 3) and move in the positive y-direction, we see that both b and c decrease, while a
increases. Both b and c have a low point at about (3, - 1.5), while a is 0 at this point. So a is defi nitely the graph of j 11
, and
one of b and c is the graph off. To see which is which, we start at the point (- 3, - 1.5) and move in the positive x-direction.
b traces out a line with negative slope, while c traces out a parabola opening downward. This tells us that b is tlle x-derivative
of c. Soc is the graph of f, b is the, graph of f ,, and a is tbe graph of f 11
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